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3 years ago

What are the three major tendons in the knee?

Physics

2 answers:

kati45 [8]3 years ago

6 0

Answer:

The two important tendons in the knee are (1) the quadriceps tendon connecting the quadriceps muscle, which lies on the front of the thigh, to the patella and (2) the patellar tendon connecting the patella to the tibia (technically, this is a ligament because it connects two bones).

Fynjy0 [20]3 years ago

3 0

Three major tendons in the knee are the quadriceps tendon patellar tendon and the hamstring tendon (not sure about the last one but I think its that) Hope this helps.

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An apple has a mass of 0.136kg and a watermelon has a mass of 9.07kg. Work out the mass of the apple as a percentage of the mass

Galina-37 [17]

7 the answer is 7 JK

3 0

3 years ago

The mass of an object is 10 kg and the velocity is 4 m/s, what is the momentum?

eduard

The answer is 40 kg. m/s.

Formula for momentum:

p=mv

p=(10 kg.)(4 m/s)

So, therefore, the final answer is p=40 kg. m/s.

I hope this helped answer your question. Enjoy your day, and take care!

3 0

3 years ago

Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0

4 years ago

Lien uses a spring scale to pull a block toward the right across the lab table. The scale reads 8 N. Which force should Lien con

lana66690 [7]

Answer:

<h2>Frictional force</h2>

Explanation:

For the block placed on the table, there are several force acting on the body along the horizontal and vertical axis. All this forces tends to keep the body in a state of equilibrium.The forces acting along the horizontal are the moving force (Fm) and the frictional force (Ff).

Frictional force are forces that acts opposite to the force that causes the body to move (moving force).

If Lien uses a spring scale to pull a block toward the right across the lab table and the scale reads 8 N, this means that the force that causes the body to move is the 8N force (moving force).

Taking the sum of force along the horizontal;

$\sum fx = ma_x$

Since the body is static, max = 0

$\sum fx = 0\\fm+(-Ff)= 0$

Note that the frictional force acts is the force of opposition acting in the negative x direction.

$fm = 0+Ff\\fm = Ff$

Since Fm = 8N, Ff will also be equal to 8N.

Based on the above proof, Lien can also conclude that 8 N is a frictional force

6 0

3 years ago

A man is flying in a hot-air balloon in a straight line at a constant rate of 4 feet per second, while maintaining it at a const

Tanya [424]

Answer: the distance between him and his friend is 218.39 feet

Explanation:

Given the data in the question;

FROM IMAGE A;

distance travelled in minute and a half

⇒ (60 + 30)sec × 4ft/sec = 360 fts

FROM IMAGE B

tan35°= h/x --- equ 1

and tan36° = h/(360 - x), so h = (360 - x)tan36°

we substitute value of h into euq 1

tan35° = (360 - x)tan36°/x

xtan35° = (360 - x)tan36°

0.7002x = 261.5553 - 0.7265x

0.7002x + 0.7265x = 261.5553

1.4267x = 261.5553

x = 183.32 feet

360 - x ⇒ ( 360 - 183.32) = 176.68

FROM IMAGE C

Let the distance between them be d

cos36° = 176.68 / d

dcos36° = 176.68

d = 176.68 / 0.809

d = 218.39 feet

Therefore the distance between him and his friend is 218.39 feet

6 0

3 years ago